Why is not every bounded function Riemann integrable?

Question

I know that not every bounded function $f$ on an interval $[a,b]$ is Riemann integrable. However, I seem to have shown that every such function is Riemann integrable. I know something must be wrong with my proof, but I can't figure out what it is. Here is what I did.

Let $M$ be a bound of $f$ on $[a,b]$. Thus $|f(x)| \leq M$ for all $x\in[a,b]$. Let $P$ and $Q$ be two arbitrary tagged partitions of $[a,b]$, and let $R(f,P)$ and $R(f,Q)$ be the corresponding Riemann sums, respectively. We have that

$\begin{align*} |R(f,P) - R(f,Q) | &= \left\vert\sum_{k=1}^{n}f(x_k^*)(x_k-x_{k-1}) - \sum_{k=1}^{n}f(y_k^*)(y_k-y_{k-1}) \right\vert\\ &\leq \sum_{k=1}^{n}|f(x_k^*)|(x_k-x_{k-1}) + \sum_{k=1}^{n}|f(y_k^*)|(y_k-y_{k-1})\\ &\leq \sum_{k=1}^{n}M(x_k-x_{k-1}) + \sum_{k=1}^{n}M(y_k-y_{k-1}). \end{align*}$

Here, $x_k^*$ and $y_k^*$ are the tags of $P$ and $Q$, respectively. Now, for any $\epsilon > 0$ let $\delta = \frac{\epsilon}{2Mn}$. Then if the norms of $P$ and $Q$ are each less than $\delta$,

\begin{align*} \sum_{k=1}^{n}M(x_k-x_{k-1}) + \sum_{k=1}^{n}M(y_k-y_{k-1}) &\leq \sum_{k=1}^{n}M\frac{\epsilon}{2Mn} + \sum_{k=1}^{n}M\frac{\epsilon}{2Mn} \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*}

This implies that the partitions satisfy the Cauchy criterion and thus $f$ is Riemann integrable.

Where am I going wrong?

Answer 1

You made the assumption that all tagged partitions have the same number of points $n$. But if you want the norm of your partition to be arbitrarily small, then $n$ will grow larger.

Answer 2

What about an indicator function, like $\chi_{\Bbb Q}$, which is given by :

$$\chi_{\Bbb Q}(x)=\begin {cases}1\,,x\in\Bbb Q\\0\,,x\notin \Bbb Q\end{cases}$$.

It's bounded. But is it integrable? Not Riemann integrable. It's Lebesgue integral is zero however. This requires some measure theory.